the value of g at the centre of the earth according to eqn. g=GM/R^2
is infinity but really its 0 why??
Answers
Answered by
1
At the centre of the earth anybody surrounds with approximately equal mass from every side . force from every/all masses being cancelled by similar force from opposite side. That is the reason why g is 0 at the centre of the earth. There is no acceleration due to gravity.
sameer500kumar:
then why not at surface of earth
Answered by
1
There are two regions to consider; the region outside the Earth and the region inside.
The solution inside (non-zero mass density) and the solution outside (zero mass density) are different but must give the same value at the radius of the surface.
Assuming a uniform mass density inside, the gravitational acceleration inside is proportional to the radius and, thus, goes to zero at the center.
The gravitational attraction from the Earth as you dig down into the Earth is due only to the mass of the Earth enclosed by the sphere defined by the radius that you are at (see the derivation of Gauss's theorem for a proof of this). Therefore, as you go further down into the Earth, the amount of mass that is pulling you towards the center decreases, and the form of the equation that you cited changes.
Outside the earth (such that R>RER>RE), as you said, the gravitational acceleration is (classically),
g=GMR2g=GMR2
But once you enter the earth (at a radius such that R<RER<RE), the mass that is gravitationally attracting you gets smaller, according to the ration of the volume of the sphere inside your radius, and the sphere inside the Earth's radius11.
M(R)=M43πR343πR3EM(R)=M43πR343πRE3
M(R)=M(RRE)3M(R)=M(RRE)3
Inserting that into the equation above, we find that the acceleration for R<RER<RE is
g=GR2M(RRE)3g=GR2M(RRE)3
Simplifying that gives us
g=GMR3ERg=GMRE3R
Since the first part of that term is a constant, we can say α=GMR3Eα=GMRE3, and
g=αR
The solution inside (non-zero mass density) and the solution outside (zero mass density) are different but must give the same value at the radius of the surface.
Assuming a uniform mass density inside, the gravitational acceleration inside is proportional to the radius and, thus, goes to zero at the center.
The gravitational attraction from the Earth as you dig down into the Earth is due only to the mass of the Earth enclosed by the sphere defined by the radius that you are at (see the derivation of Gauss's theorem for a proof of this). Therefore, as you go further down into the Earth, the amount of mass that is pulling you towards the center decreases, and the form of the equation that you cited changes.
Outside the earth (such that R>RER>RE), as you said, the gravitational acceleration is (classically),
g=GMR2g=GMR2
But once you enter the earth (at a radius such that R<RER<RE), the mass that is gravitationally attracting you gets smaller, according to the ration of the volume of the sphere inside your radius, and the sphere inside the Earth's radius11.
M(R)=M43πR343πR3EM(R)=M43πR343πRE3
M(R)=M(RRE)3M(R)=M(RRE)3
Inserting that into the equation above, we find that the acceleration for R<RER<RE is
g=GR2M(RRE)3g=GR2M(RRE)3
Simplifying that gives us
g=GMR3ERg=GMRE3R
Since the first part of that term is a constant, we can say α=GMR3Eα=GMRE3, and
g=αR
Attachments:
Similar questions